Solving equations involving square roots












4












$begingroup$


I am a student and I often encounter these type of equations:



$$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6$$



I usually solve these by taking one term ($sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.



Thanks






algebra-precalculus







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$endgroup$












  • $begingroup$
    What do you want to solve for? For a variable, or in general a parametric form for all possible solutions?
    $endgroup$
    – KKZiomek
    Feb 2 at 14:56










  • $begingroup$
    I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 14:59










  • $begingroup$
    All possible solutions
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 15:02






  • 1




    $begingroup$
    i might be wrong, but that is known as ellipse
    $endgroup$
    – aaaaaa
    Feb 2 at 18:18
















4












$begingroup$


I am a student and I often encounter these type of equations:



$$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6$$



I usually solve these by taking one term ($sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.



Thanks






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you want to solve for? For a variable, or in general a parametric form for all possible solutions?
    $endgroup$
    – KKZiomek
    Feb 2 at 14:56










  • $begingroup$
    I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 14:59










  • $begingroup$
    All possible solutions
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 15:02






  • 1




    $begingroup$
    i might be wrong, but that is known as ellipse
    $endgroup$
    – aaaaaa
    Feb 2 at 18:18














4












4








4


1



$begingroup$


I am a student and I often encounter these type of equations:



$$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6$$



I usually solve these by taking one term ($sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.



Thanks






algebra-precalculus







share|cite|improve this question











$endgroup$




I am a student and I often encounter these type of equations:



$$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6$$



I usually solve these by taking one term ($sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.



Thanks






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 14:42









caverac

14.8k31130




14.8k31130










asked Feb 2 at 14:31









Shantanu KaushikShantanu Kaushik

443




443












  • $begingroup$
    What do you want to solve for? For a variable, or in general a parametric form for all possible solutions?
    $endgroup$
    – KKZiomek
    Feb 2 at 14:56










  • $begingroup$
    I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 14:59










  • $begingroup$
    All possible solutions
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 15:02






  • 1




    $begingroup$
    i might be wrong, but that is known as ellipse
    $endgroup$
    – aaaaaa
    Feb 2 at 18:18


















  • $begingroup$
    What do you want to solve for? For a variable, or in general a parametric form for all possible solutions?
    $endgroup$
    – KKZiomek
    Feb 2 at 14:56










  • $begingroup$
    I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 14:59










  • $begingroup$
    All possible solutions
    $endgroup$
    – Shantanu Kaushik
    Feb 2 at 15:02






  • 1




    $begingroup$
    i might be wrong, but that is known as ellipse
    $endgroup$
    – aaaaaa
    Feb 2 at 18:18
















$begingroup$
What do you want to solve for? For a variable, or in general a parametric form for all possible solutions?
$endgroup$
– KKZiomek
Feb 2 at 14:56




$begingroup$
What do you want to solve for? For a variable, or in general a parametric form for all possible solutions?
$endgroup$
– KKZiomek
Feb 2 at 14:56












$begingroup$
I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6
$endgroup$
– Shantanu Kaushik
Feb 2 at 14:59




$begingroup$
I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6
$endgroup$
– Shantanu Kaushik
Feb 2 at 14:59












$begingroup$
All possible solutions
$endgroup$
– Shantanu Kaushik
Feb 2 at 15:02




$begingroup$
All possible solutions
$endgroup$
– Shantanu Kaushik
Feb 2 at 15:02




1




1




$begingroup$
i might be wrong, but that is known as ellipse
$endgroup$
– aaaaaa
Feb 2 at 18:18




$begingroup$
i might be wrong, but that is known as ellipse
$endgroup$
– aaaaaa
Feb 2 at 18:18










4 Answers
4






active

oldest

votes


















2












$begingroup$

A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.



Given that
$$begin{align}sqrt{x^2 +(y-2)^2} + sqrt{x^2+(y+2)^2} &= 6 &[1]\ end{align}$$
and that
$$begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\ end{align}$$



we have (by $frac{[2]}{[1]}$)
$$begin{align}sqrt{x^2 +(y-2)^2} - sqrt{x^2+(y+2)^2} = frac{-8y}{6} &= -frac{4}{3}y &[3]\ end{align}$$



Adding $[1]$ and $[3]$ gives us
$$begin{align}
2sqrt{x^2 +(y-2)^2} &= 6-frac{4}{3}y &[4]\
4(x^2 +(y-2)^2) &= bigg(6-frac{4}{3}ybigg)^2\
4x^2 +4y^2-16y+16 &= frac{16}{9}y^2-16y+36\
x^2 +y^2+4 &= frac{4}{9}y^2+9\
x^2 +frac{5}{9}y^2 &= 5\
frac{x^2}{5} +frac{y^2}{9} &= 1\
end{align}$$






share|cite|improve this answer









$endgroup$





















    8












    $begingroup$

    The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(sqrt 5, 0)$, $(-sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by:
    $$frac{x^2}{5}+frac{y^2}{9} = 1$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
      $endgroup$
      – Peter Foreman
      Feb 2 at 15:13






    • 1




      $begingroup$
      You can simply edit your answer.
      $endgroup$
      – Yves Daoust
      Feb 2 at 15:17










    • $begingroup$
      Thank you kind sir
      $endgroup$
      – Shantanu Kaushik
      Feb 2 at 15:33



















    4












    $begingroup$

    Use



    $$sqrt a+sqrt b=c$$
    then



    $$a+2sqrt{ab}+b=c^2$$



    then



    $$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$



    and finally



    $$c^4-2c^2(a+b)+(a-b)^2=0.$$




    In your case,



    $$1296-144(x^2+y^2+4)+64y^2=0$$



    or



    $$left(frac x{sqrt5}right)^2+left(frac y3right)^2=1,$$ which is a centered, axis-aligned ellipse.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you very much sir
      $endgroup$
      – Shantanu Kaushik
      Feb 2 at 15:34



















    3












    $begingroup$

    Denote:
    $$begin{cases}x^2 + (y-2)^2=t^2 \ x^2+(y+2)^2=t^2+8yend{cases}.$$
    Then:
    $$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6 Rightarrow \
    t+sqrt{t^2+8y}=6 Rightarrow \
    t^2+8y=36-12t+t^2 Rightarrow \
    t=frac{9-2y}{3}.$$
    Plug it into the first equation:
    $$x^2+(y-2)^2=left(frac{9-2y}{3}right)^2 Rightarrow \
    9x^2+9y^2-36y+36=81-36y+4y^2 Rightarrow \
    9x^2+5y^2=45 Rightarrow \
    frac{x^2}{5}+frac{y^2}{9}=1.$$






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.



      Given that
      $$begin{align}sqrt{x^2 +(y-2)^2} + sqrt{x^2+(y+2)^2} &= 6 &[1]\ end{align}$$
      and that
      $$begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\ end{align}$$



      we have (by $frac{[2]}{[1]}$)
      $$begin{align}sqrt{x^2 +(y-2)^2} - sqrt{x^2+(y+2)^2} = frac{-8y}{6} &= -frac{4}{3}y &[3]\ end{align}$$



      Adding $[1]$ and $[3]$ gives us
      $$begin{align}
      2sqrt{x^2 +(y-2)^2} &= 6-frac{4}{3}y &[4]\
      4(x^2 +(y-2)^2) &= bigg(6-frac{4}{3}ybigg)^2\
      4x^2 +4y^2-16y+16 &= frac{16}{9}y^2-16y+36\
      x^2 +y^2+4 &= frac{4}{9}y^2+9\
      x^2 +frac{5}{9}y^2 &= 5\
      frac{x^2}{5} +frac{y^2}{9} &= 1\
      end{align}$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.



        Given that
        $$begin{align}sqrt{x^2 +(y-2)^2} + sqrt{x^2+(y+2)^2} &= 6 &[1]\ end{align}$$
        and that
        $$begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\ end{align}$$



        we have (by $frac{[2]}{[1]}$)
        $$begin{align}sqrt{x^2 +(y-2)^2} - sqrt{x^2+(y+2)^2} = frac{-8y}{6} &= -frac{4}{3}y &[3]\ end{align}$$



        Adding $[1]$ and $[3]$ gives us
        $$begin{align}
        2sqrt{x^2 +(y-2)^2} &= 6-frac{4}{3}y &[4]\
        4(x^2 +(y-2)^2) &= bigg(6-frac{4}{3}ybigg)^2\
        4x^2 +4y^2-16y+16 &= frac{16}{9}y^2-16y+36\
        x^2 +y^2+4 &= frac{4}{9}y^2+9\
        x^2 +frac{5}{9}y^2 &= 5\
        frac{x^2}{5} +frac{y^2}{9} &= 1\
        end{align}$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.



          Given that
          $$begin{align}sqrt{x^2 +(y-2)^2} + sqrt{x^2+(y+2)^2} &= 6 &[1]\ end{align}$$
          and that
          $$begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\ end{align}$$



          we have (by $frac{[2]}{[1]}$)
          $$begin{align}sqrt{x^2 +(y-2)^2} - sqrt{x^2+(y+2)^2} = frac{-8y}{6} &= -frac{4}{3}y &[3]\ end{align}$$



          Adding $[1]$ and $[3]$ gives us
          $$begin{align}
          2sqrt{x^2 +(y-2)^2} &= 6-frac{4}{3}y &[4]\
          4(x^2 +(y-2)^2) &= bigg(6-frac{4}{3}ybigg)^2\
          4x^2 +4y^2-16y+16 &= frac{16}{9}y^2-16y+36\
          x^2 +y^2+4 &= frac{4}{9}y^2+9\
          x^2 +frac{5}{9}y^2 &= 5\
          frac{x^2}{5} +frac{y^2}{9} &= 1\
          end{align}$$






          share|cite|improve this answer









          $endgroup$



          A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.



          Given that
          $$begin{align}sqrt{x^2 +(y-2)^2} + sqrt{x^2+(y+2)^2} &= 6 &[1]\ end{align}$$
          and that
          $$begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\ end{align}$$



          we have (by $frac{[2]}{[1]}$)
          $$begin{align}sqrt{x^2 +(y-2)^2} - sqrt{x^2+(y+2)^2} = frac{-8y}{6} &= -frac{4}{3}y &[3]\ end{align}$$



          Adding $[1]$ and $[3]$ gives us
          $$begin{align}
          2sqrt{x^2 +(y-2)^2} &= 6-frac{4}{3}y &[4]\
          4(x^2 +(y-2)^2) &= bigg(6-frac{4}{3}ybigg)^2\
          4x^2 +4y^2-16y+16 &= frac{16}{9}y^2-16y+36\
          x^2 +y^2+4 &= frac{4}{9}y^2+9\
          x^2 +frac{5}{9}y^2 &= 5\
          frac{x^2}{5} +frac{y^2}{9} &= 1\
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 2 at 23:22









          John JoyJohn Joy

          6,30911827




          6,30911827























              8












              $begingroup$

              The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(sqrt 5, 0)$, $(-sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by:
              $$frac{x^2}{5}+frac{y^2}{9} = 1$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
                $endgroup$
                – Peter Foreman
                Feb 2 at 15:13






              • 1




                $begingroup$
                You can simply edit your answer.
                $endgroup$
                – Yves Daoust
                Feb 2 at 15:17










              • $begingroup$
                Thank you kind sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:33
















              8












              $begingroup$

              The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(sqrt 5, 0)$, $(-sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by:
              $$frac{x^2}{5}+frac{y^2}{9} = 1$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
                $endgroup$
                – Peter Foreman
                Feb 2 at 15:13






              • 1




                $begingroup$
                You can simply edit your answer.
                $endgroup$
                – Yves Daoust
                Feb 2 at 15:17










              • $begingroup$
                Thank you kind sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:33














              8












              8








              8





              $begingroup$

              The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(sqrt 5, 0)$, $(-sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by:
              $$frac{x^2}{5}+frac{y^2}{9} = 1$$






              share|cite|improve this answer











              $endgroup$



              The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(sqrt 5, 0)$, $(-sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by:
              $$frac{x^2}{5}+frac{y^2}{9} = 1$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 2 at 19:00









              S. Sharma

              31110




              31110










              answered Feb 2 at 15:05









              Peter ForemanPeter Foreman

              7,2371318




              7,2371318












              • $begingroup$
                I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
                $endgroup$
                – Peter Foreman
                Feb 2 at 15:13






              • 1




                $begingroup$
                You can simply edit your answer.
                $endgroup$
                – Yves Daoust
                Feb 2 at 15:17










              • $begingroup$
                Thank you kind sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:33


















              • $begingroup$
                I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
                $endgroup$
                – Peter Foreman
                Feb 2 at 15:13






              • 1




                $begingroup$
                You can simply edit your answer.
                $endgroup$
                – Yves Daoust
                Feb 2 at 15:17










              • $begingroup$
                Thank you kind sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:33
















              $begingroup$
              I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
              $endgroup$
              – Peter Foreman
              Feb 2 at 15:13




              $begingroup$
              I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case.
              $endgroup$
              – Peter Foreman
              Feb 2 at 15:13




              1




              1




              $begingroup$
              You can simply edit your answer.
              $endgroup$
              – Yves Daoust
              Feb 2 at 15:17




              $begingroup$
              You can simply edit your answer.
              $endgroup$
              – Yves Daoust
              Feb 2 at 15:17












              $begingroup$
              Thank you kind sir
              $endgroup$
              – Shantanu Kaushik
              Feb 2 at 15:33




              $begingroup$
              Thank you kind sir
              $endgroup$
              – Shantanu Kaushik
              Feb 2 at 15:33











              4












              $begingroup$

              Use



              $$sqrt a+sqrt b=c$$
              then



              $$a+2sqrt{ab}+b=c^2$$



              then



              $$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$



              and finally



              $$c^4-2c^2(a+b)+(a-b)^2=0.$$




              In your case,



              $$1296-144(x^2+y^2+4)+64y^2=0$$



              or



              $$left(frac x{sqrt5}right)^2+left(frac y3right)^2=1,$$ which is a centered, axis-aligned ellipse.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thank you very much sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:34
















              4












              $begingroup$

              Use



              $$sqrt a+sqrt b=c$$
              then



              $$a+2sqrt{ab}+b=c^2$$



              then



              $$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$



              and finally



              $$c^4-2c^2(a+b)+(a-b)^2=0.$$




              In your case,



              $$1296-144(x^2+y^2+4)+64y^2=0$$



              or



              $$left(frac x{sqrt5}right)^2+left(frac y3right)^2=1,$$ which is a centered, axis-aligned ellipse.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thank you very much sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:34














              4












              4








              4





              $begingroup$

              Use



              $$sqrt a+sqrt b=c$$
              then



              $$a+2sqrt{ab}+b=c^2$$



              then



              $$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$



              and finally



              $$c^4-2c^2(a+b)+(a-b)^2=0.$$




              In your case,



              $$1296-144(x^2+y^2+4)+64y^2=0$$



              or



              $$left(frac x{sqrt5}right)^2+left(frac y3right)^2=1,$$ which is a centered, axis-aligned ellipse.






              share|cite|improve this answer











              $endgroup$



              Use



              $$sqrt a+sqrt b=c$$
              then



              $$a+2sqrt{ab}+b=c^2$$



              then



              $$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$



              and finally



              $$c^4-2c^2(a+b)+(a-b)^2=0.$$




              In your case,



              $$1296-144(x^2+y^2+4)+64y^2=0$$



              or



              $$left(frac x{sqrt5}right)^2+left(frac y3right)^2=1,$$ which is a centered, axis-aligned ellipse.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 2 at 15:14

























              answered Feb 2 at 15:09









              Yves DaoustYves Daoust

              133k676231




              133k676231












              • $begingroup$
                Thank you very much sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:34


















              • $begingroup$
                Thank you very much sir
                $endgroup$
                – Shantanu Kaushik
                Feb 2 at 15:34
















              $begingroup$
              Thank you very much sir
              $endgroup$
              – Shantanu Kaushik
              Feb 2 at 15:34




              $begingroup$
              Thank you very much sir
              $endgroup$
              – Shantanu Kaushik
              Feb 2 at 15:34











              3












              $begingroup$

              Denote:
              $$begin{cases}x^2 + (y-2)^2=t^2 \ x^2+(y+2)^2=t^2+8yend{cases}.$$
              Then:
              $$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6 Rightarrow \
              t+sqrt{t^2+8y}=6 Rightarrow \
              t^2+8y=36-12t+t^2 Rightarrow \
              t=frac{9-2y}{3}.$$
              Plug it into the first equation:
              $$x^2+(y-2)^2=left(frac{9-2y}{3}right)^2 Rightarrow \
              9x^2+9y^2-36y+36=81-36y+4y^2 Rightarrow \
              9x^2+5y^2=45 Rightarrow \
              frac{x^2}{5}+frac{y^2}{9}=1.$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Denote:
                $$begin{cases}x^2 + (y-2)^2=t^2 \ x^2+(y+2)^2=t^2+8yend{cases}.$$
                Then:
                $$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6 Rightarrow \
                t+sqrt{t^2+8y}=6 Rightarrow \
                t^2+8y=36-12t+t^2 Rightarrow \
                t=frac{9-2y}{3}.$$
                Plug it into the first equation:
                $$x^2+(y-2)^2=left(frac{9-2y}{3}right)^2 Rightarrow \
                9x^2+9y^2-36y+36=81-36y+4y^2 Rightarrow \
                9x^2+5y^2=45 Rightarrow \
                frac{x^2}{5}+frac{y^2}{9}=1.$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Denote:
                  $$begin{cases}x^2 + (y-2)^2=t^2 \ x^2+(y+2)^2=t^2+8yend{cases}.$$
                  Then:
                  $$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6 Rightarrow \
                  t+sqrt{t^2+8y}=6 Rightarrow \
                  t^2+8y=36-12t+t^2 Rightarrow \
                  t=frac{9-2y}{3}.$$
                  Plug it into the first equation:
                  $$x^2+(y-2)^2=left(frac{9-2y}{3}right)^2 Rightarrow \
                  9x^2+9y^2-36y+36=81-36y+4y^2 Rightarrow \
                  9x^2+5y^2=45 Rightarrow \
                  frac{x^2}{5}+frac{y^2}{9}=1.$$






                  share|cite|improve this answer









                  $endgroup$



                  Denote:
                  $$begin{cases}x^2 + (y-2)^2=t^2 \ x^2+(y+2)^2=t^2+8yend{cases}.$$
                  Then:
                  $$sqrt{x^2 + (y-2)^2} + sqrt{x^2 + (y+2)^2} = 6 Rightarrow \
                  t+sqrt{t^2+8y}=6 Rightarrow \
                  t^2+8y=36-12t+t^2 Rightarrow \
                  t=frac{9-2y}{3}.$$
                  Plug it into the first equation:
                  $$x^2+(y-2)^2=left(frac{9-2y}{3}right)^2 Rightarrow \
                  9x^2+9y^2-36y+36=81-36y+4y^2 Rightarrow \
                  9x^2+5y^2=45 Rightarrow \
                  frac{x^2}{5}+frac{y^2}{9}=1.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 2 at 16:27









                  farruhotafarruhota

                  22.1k2942




                  22.1k2942






























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